3. 3
Unevaluated Terms
Prolog operators allow terms to be written
more concisely, but are not evaluated
These are all the same Prolog term:
That term does not unify with 7
+(1,*(2,3))
1+ *(2,3)
+(1,2*3)
(1+(2*3))
1+2*3
4. 4
Evaluating Expressions
The predefined predicate is can be used to
evaluate a term that is a numeric expression
is(X,Y) evaluates the term Y and unifies
X with the resulting atom
It is usually used as an operator
?- X is 1+2*3.
X = 7
Yes
5. 5
Instantiation Is Required
?- Y=X+2, X=1.
Y = 1+2
X = 1
Yes
?- Y is X+2, X=1.
ERROR: Arguments are not sufficiently instantiated
?- X=1, Y is X+2.
X = 1
Y = 3
Yes
6. 6
Evaluable Predicates
For X is Y, the predicates that appear in
Y have to be evaluable predicates
This includes things like the predefined
operators +, -, * and /
There are also other predefined evaluable
predicates, like abs(Z) and sqrt(Z)
7. 7
Real Values And Integers
?- X is 1/2.
X = 0.5
Yes
?- X is 1.0/2.0.
X = 0.5
Yes
?- X is 2/1.
X = 2
Yes
?- X is 2.0/1.0.
X = 2
Yes
There are two numeric types:
integer and real.
Most of the evaluable
predicates are overloaded for
all combinations.
Prolog is dynamically typed;
the types are used at runtime
to resolve the overloading.
But note that the goal 2=2.0
would fail.
8. 8
Comparisons
Numeric comparison operators:
<, >, =<, >=, =:=, ==
To solve a numeric comparison goal, Prolog
evaluates both sides and compares the
results numerically
So both sides must be fully instantiated
9. 9
Comparisons
?- 1+2 < 1*2.
No
?- 1<2.
Yes
?- 1+2>=1+3.
No
?- X is 1-3, Y is 0-2, X =:= Y.
X = -2
Y = -2
Yes
10. 10
Equalities In Prolog
We have used three different but related
equality operators:
– X is Y evaluates Y and unifies the result with X:
3 is 1+2 succeeds, but 1+2 is 3 fails
– X = Y unifies X and Y, with no evaluation: both
3 = 1+2 and 1+2 = 3 fail
– X =:= Y evaluates both and compares: both
3 =:= 1+2 and 1+2 =:= 3 succeed
Any evaluated term must be fully instantiated
14. 14
Example: gcd
gcd(X,Y,Z) :-
X =:= Y,
Z is X.
gcd(X,Y,Denom) :-
X < Y,
NewY is Y - X,
gcd(X,NewY,Denom).
gcd(X,Y,Denom) :-
X > Y,
NewX is X - Y,
gcd(NewX,Y,Denom).
Note: not just
gcd(X,X,X)
15. 15
The gcd Predicate At Work
?- gcd(5,5,X).
X = 5
Yes
?- gcd(12,21,X).
X = 3
Yes
?- gcd(91,105,X).
X = 7
Yes
?- gcd(91,X,7).
ERROR: Arguments are not sufficiently instantiated
16. 16
Example: factorial
factorial(X,1) :-
X =:= 1.
factorial(X,Fact) :-
X > 1,
NewX is X - 1,
factorial(NewX,NF),
Fact is X * NF.
?- factorial(5,X).
X = 120
Yes
?- factorial(20,X).
X = 2.4329e+018
Yes
?- factorial(-2,X).
No
18. 18
Problem Space Search
Prolog’s strength is (obviously) not numeric
computation
The kinds of problems it does best on are
those that involve problem space search
– You give a logical definition of the solution
– Then let Prolog find it
19. 19
The Knapsack Problem
You are packing for a camping trip
Your pantry contains these items:
Your knapsack holds 4 kg.
What choice <= 4 kg. maximizes calories?
Item Weight in kilograms Calories
bread 4 9200
pasta 2 4600
peanut butter 1 6700
baby food 3 6900
20. 20
Greedy Methods Do Not Work
Most calories first: bread only, 9200
Lightest first: peanut butter + pasta, 11300
(Best choice: peanut butter + baby food,
13600)
Item Weight in kilograms Calories
bread 4 9200
pasta 2 4600
peanut butter 1 6700
baby food 3 6900
21. 21
Search
No algorithm for this problem is known that
– Always gives the best answer, and
– Takes less than exponential time
So brute-force search is nothing to be
ashamed of here
That’s good, since search is something
Prolog does really well
22. 22
Representation
We will represent each food item as a term
food(N,W,C)
Pantry in our example is
[food(bread,4,9200),
food(pasta,2,4500),
food(peanutButter,1,6700),
food(babyFood,3,6900)]
Same representation for knapsack contents
23. 23
/*
weight(L,N) takes a list L of food terms, each
of the form food(Name,Weight,Calories). We
unify N with the sum of all the Weights.
*/
weight([],0).
weight([food(_,W,_) | Rest], X) :-
weight(Rest,RestW),
X is W + RestW.
/*
calories(L,N) takes a list L of food terms, each
of the form food(Name,Weight,Calories). We
unify N with the sum of all the Calories.
*/
calories([],0).
calories([food(_,_,C) | Rest], X) :-
calories(Rest,RestC),
X is C + RestC.
24. 24
/*
subseq(X,Y) succeeds when list X is the same as
list Y, but with zero or more elements omitted.
This can be used with any pattern of instantiations.
*/
subseq([],[]).
subseq([Item | RestX], [Item | RestY]) :-
subseq(RestX,RestY).
subseq(X, [_ | RestY]) :-
subseq(X,RestY).
A subsequence of a list is a copy of the list
with any number of elements omitted
(Knapsacks are subsequences of the pantry)
25. 25
?- subseq([1,3],[1,2,3,4]).
Yes
?- subseq(X,[1,2,3]).
X = [1, 2, 3] ;
X = [1, 2] ;
X = [1, 3] ;
X = [1] ;
X = [2, 3] ;
X = [2] ;
X = [3] ;
X = [] ;
No
Note that subseq can do more
than just test whether one list is a
subsequence of another; it can
generate subsequences, which is
how we will use it for the
knapsack problem.
26. 26
/*
knapsackDecision(Pantry,Capacity,Goal,Knapsack) takes
a list Pantry of food terms, a positive number
Capacity, and a positive number Goal. We unify
Knapsack with a subsequence of Pantry representing
a knapsack with total calories >= goal, subject to
the constraint that the total weight is =< Capacity.
*/
knapsackDecision(Pantry,Capacity,Goal,Knapsack) :-
subseq(Knapsack,Pantry),
weight(Knapsack,Weight),
Weight =< Capacity,
calories(Knapsack,Calories),
Calories >= Goal.
27. 27
This decides whether there is a solution that
meets the given calorie goal
Not exactly the answer we want…
?- knapsackDecision(
| [food(bread,4,9200),
| food(pasta,2,4500),
| food(peanutButter,1,6700),
| food(babyFood,3,6900)],
| 4,
| 10000,
| X).
X = [food(pasta, 2, 4500),
food(peanutButter, 1, 6700)]
Yes
28. 28
Decision And Optimization
We solved the knapsack decision problem
What we wanted to solve was the knapsack
optimization problem
To do that, we will use another predefined
predicate: findall
29. 29
The findall Predicate
findall(X,Goal,L)
– Finds all the ways of proving Goal
– For each, applies to X the same substitution that
made a provable instance of Goal
– Unifies L with the list of all those X’s
30. 30
Counting The Solutions
This shows there were four ways of proving
subseq(_,[1,2])
Collected a list of 1’s, one for each proof
?- findall(1,subseq(_,[1,2]),L).
L = [1, 1, 1, 1]
Yes
31. 31
Collecting The Instances
The first and second parameters to
findall are the same
This collects all four provable instances of
the goal subseq(X,[1,2])
?- findall(subseq(X,[1,2]),subseq(X,[1,2]),L).
X = _G396
L = [subseq([1, 2], [1, 2]), subseq([1], [1, 2]),
subseq([2], [1, 2]), subseq([], [1, 2])]
Yes
32. 32
Collecting Particular
Substitutions
A common use of findall: the first
parameter is a variable from the second
This collects all four X’s that make the goal
subseq(X,[1,2]) provable
?- findall(X,subseq(X,[1,2]),L).
X = _G312
L = [[1, 2], [1], [2], []]
Yes
33. 33
/*
legalKnapsack(Pantry,Capacity,Knapsack) takes a list
Pantry of food terms and a positive number Capacity.
We unify Knapsack with a subsequence of Pantry whose
total weight is =< Capacity.
*/
legalKnapsack(Pantry,Capacity,Knapsack):-
subseq(Knapsack,Pantry),
weight(Knapsack,W),
W =< Capacity.
34. 34
/*
maxCalories(List,Result) takes a List of lists of
food terms. We unify Result with an element from the
list that maximizes the total calories. We use a
helper predicate maxC that takes four paramters: the
remaining list of lists of food terms, the best list
of food terms seen so far, its total calories, and
the final result.
*/
maxC([],Sofar,_,Sofar).
maxC([First | Rest],_,MC,Result) :-
calories(First,FirstC),
MC =< FirstC,
maxC(Rest,First,FirstC,Result).
maxC([First | Rest],Sofar,MC,Result) :-
calories(First,FirstC),
MC > FirstC,
maxC(Rest,Sofar,MC,Result).
maxCalories([First | Rest],Result) :-
calories(First,FirstC),
maxC(Rest,First,FirstC,Result).
35. 35
/*
knapsackOptimization(Pantry,Capacity,Knapsack) takes
a list Pantry of food items and a positive integer
Capacity. We unify Knapsack with a subsequence of
Pantry representing a knapsack of maximum total
calories, subject to the constraint that the total
weight is =< Capacity.
*/
knapsackOptimization(Pantry,Capacity,Knapsack) :-
findall(K,legalKnapsack(Pantry,Capacity,K),L),
maxCalories(L,Knapsack).
38. 38
The 8-Queens Problem
Chess background:
– Played on an 8-by-8 grid
– Queen can move any number of spaces
vertically, horizontally or diagonally
– Two queens are in check if they are in the same
row, column or diagonal, so that one could
move to the other’s square
The problem: place 8 queens on an empty
chess board so that no queen is in check
39. 39
Representation
We could represent a queen in column 2,
row 5 with the term queen(2,5)
But it will be more readable if we use
something more compact
Since there will be no other pieces—no
pawn(X,Y) or king(X,Y)—we will just
use a term of the form X/Y
(We won’t evaluate it as a quotient)
41. 41
/*
nocheck(X/Y,L) takes a queen X/Y and a list
of queens. We succeed if and only if the X/Y
queen holds none of the others in check.
*/
nocheck(_, []).
nocheck(X/Y, [X1/Y1 | Rest]) :-
X == X1,
Y == Y1,
abs(Y1-Y) == abs(X1-X),
nocheck(X/Y, Rest).
42. 42
/*
legal(L) succeeds if L is a legal placement of
queens: all coordinates in range and no queen
in check.
*/
legal([]).
legal([X/Y | Rest]) :-
legal(Rest),
member(X,[1,2,3,4,5,6,7,8]),
member(Y,[1,2,3,4,5,6,7,8]),
nocheck(X/Y, Rest).
43. 43
Adequate
This is already enough to solve the
problem: the query legal(X) will find all
legal configurations:
?- legal(X).
X = [] ;
X = [1/1] ;
X = [1/2] ;
X = [1/3]
44. 44
8-Queens Solution
Of course that will take too long: it finds all
64 legal 1-queens solutions, then starts on
the 2-queens solutions, and so on
To make it concentrate right away on
8-queens, we can give a different query:
?- X = [_,_,_,_,_,_,_,_], legal(X).
X = [8/4, 7/2, 6/7, 5/3, 4/6, 3/8, 2/5, 1/1]
Yes
46. 46
Room For Improvement
Slow
Finds trivial permutations after the first:
?- X = [_,_,_,_,_,_,_,_], legal(X).
X = [8/4, 7/2, 6/7, 5/3, 4/6, 3/8, 2/5, 1/1] ;
X = [7/2, 8/4, 6/7, 5/3, 4/6, 3/8, 2/5, 1/1] ;
X = [8/4, 6/7, 7/2, 5/3, 4/6, 3/8, 2/5, 1/1] ;
X = [6/7, 8/4, 7/2, 5/3, 4/6, 3/8, 2/5, 1/1]
47. 47
An Improvement
Clearly every solution has 1 queen in each
column
So every solution can be written in a fixed
order, like this:
X=[1/_,2/_,3/_,4/_,5/_,6/_,7/_,8/_]
Starting with a goal term of that form will
restrict the search (speeding it up) and avoid
those trivial permutations
48. 48
/*
eightqueens(X) succeeds if X is a legal
placement of eight queens, listed in order
of their X coordinates.
*/
eightqueens(X) :-
X = [1/_,2/_,3/_,4/_,5/_,6/_,7/_,8/_],
legal(X).
49. 49
nocheck(_, []).
nocheck(X/Y, [X1/Y1 | Rest]) :-
% X == X1, assume the X's are distinct
Y == Y1,
abs(Y1-Y) == abs(X1-X),
nocheck(X/Y, Rest).
legal([]).
legal([X/Y | Rest]) :-
legal(Rest),
% member(X,[1,2,3,4,5,6,7,8]), assume X in range
member(Y,[1,2,3,4,5,6,7,8]),
nocheck(X/Y, Rest).
Since all X-coordinates are already known
to be in range and distinct, these can be
optimized a little
50. 50
Improved 8-Queens Solution
Now much faster
Does not bother with permutations
?- eightqueens(X).
X = [1/4, 2/2, 3/7, 4/3, 5/6, 6/8, 7/5, 8/1] ;
X = [1/5, 2/2, 3/4, 4/7, 5/3, 6/8, 7/6, 8/1] ;
51. 51
An Experiment
Fails: “arguments not sufficiently
instantiated”
The member condition does not just test
in-range coordinates; it generates them
legal([]).
legal([X/Y | Rest]) :-
legal(Rest),
% member(X,[1,2,3,4,5,6,7,8]), assume X in range
1=<Y, Y=<8, % was member(Y,[1,2,3,4,5,6,7,8]),
nocheck(X/Y, Rest).
52. 52
Another Experiment
Fails: “arguments not sufficiently
instantiated”
The legal(Rest) condition must come
first, because it generates the partial
solution tested by nocheck
legal([]).
legal([X/Y | Rest]) :-
% member(X,[1,2,3,4,5,6,7,8]), assume X in range
member(Y,[1,2,3,4,5,6,7,8]),
nocheck(X/Y, Rest),
legal(Rest). % formerly the first condition
54. 54
Parts We Skipped
The cut (!)
– A goal that always succeeds, but only once
– Used to control backtracking
Exception handling
– System-generated or user-generated exceptions
– throw and catch predicates
The API
– A small ISO API; most systems provide more
– Many public Prolog libraries: network and file
I/O, graphical user interfaces, etc.
55. 55
A Small Language
We did not have to skip as much of Prolog
as we did of ML and Java
Prolog is a small language
Yet it is powerful and not easy to master
The most important things we skipped are
the techniques Prolog programmers use to
get the most out of it
